Question

A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium. The concentration of the equilibrium shows concentration of hydrogen iodude as 1.87×10^-3M.Calculate kc at this temperature

Answer 1

Kc = 50.5

Step-by-step explanation:

We determine the reaction:

H₂ + I₂ ⇄ 2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

H₂ + I₂ ⇄ 2HI

In: 0.001 0.002 -

R: x x 2x

Eq: 0.001-x 0.002-x 0.00187

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is = (HI)² / (H₂) . (I₂)

0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5